Indian J. Pure Appl. Math., 47(3): 471-489, September 2016 °c Indian National Science Academy DOI: 10.1007/s13226-016-0179-2 CONNECTION BETWEEN COMMUTATIVE ALGEBRA AND TOPOLOGY1 Sumit Kumar Upadhyay¤, Shiv Datt Kumar¤ and Raja Sridharan¤¤ ¤Department of Mathematics, Motilal Nehru National Institute of Technology Allahabad, U. P., India ¤¤School of Mathematics, Tata Institute of Fundamental Research, Colaba, Mumbai, India e-mails: [email protected], [email protected], [email protected] (Received 26 June 2014; after final revision 10 July 2015; accepted 15 December 2015) The main aim of this paper is to show how commutative algebra is connected to topology. We give underlying topological idea of some results on completable unimodular rows. Key words : Unimodular row; matrix; homotopy. 1. INTRODUCTION n Let k be a field and ~x = (x1; x2 ¢ ¢ ¢ ; xn) 2 k be a non-zero vector, then ~x can be completed to a basis of kn. We wish to have an analogue of the above statement for rings. Let A be a ring and ~a 2 An;~a 6= 0. n Then there is a natural question when can ~a be completed to a basis of A ? Suppose ~a = (a1; a2; ¢ ¢ ¢ ; an) can be completed to a basis of An. Consider these basis vectors as columns of a matrix ®, whose first n column is (a1; a2; ¢ ¢ ¢ ; an). The fact that these vectors span A imply that there exists a n £ n invertible matrix ¯ such that ®¯ = In. Conversely if there exists an invertible matrix ® 2 Mn(A) with first column n (a1; a2; ¢ ¢ ¢ ; an), then (a1; a2; ¢ ¢ ¢ ; an) can be completed to a basis of A . Since any completable row (a1; a2; ¢ ¢ ¢ ; an) is unimodular, this leads to the following problem : n Suppose (a1; a2; ¢ ¢ ¢ ; an) 2 A is a unimodular row. Then can one complete the row (a1; a2; ¢ ¢ ¢ ; an) to a matrix belonging to GLn(A)? In general answer of this question is negative. Surprisingly this is related to topology. 1This work is supported by Council of Scientific and Industrial Research grant. 472 SUMIT KUMAR UPADHYAY et al. t Suppose one can find a matrix ® 2 GLn(A) having first column (a1; a2; ¢ ¢ ¢ ; an). Then e1® = n (a1; a2; ¢ ¢ ¢ ; an), where e1 = (1; 0; ¢ ¢ ¢ ; 0). The group GLn(A) acts on A via matrix multiplication. The row (a1; a2; ¢ ¢ ¢ ; an) can be completed to a matrix in GLn(A) if and only if (a1; a2; ¢ ¢ ¢ ; an) lies in the orbit of (1; 0; ¢ ¢ ¢ ; 0) under the GLn(A) action (a similar statement holds for SLn(A)). In this paper ring means commutative ring with identity and (a1; a2; ¢ ¢ ¢ ; an) » (b1; b2; ¢ ¢ ¢ ; bn) means 9 ® 2 GLn(A) such that (a1; a2; ¢ ¢ ¢ ; an)® = (b1; b2; ¢ ¢ ¢ ; bn). 2 Example 1.1 : Let (x1; x2) 2 Z . Then (x1; x2) is unimodular if and onlyà if x1; x2 are! relatively prime. x1 ¡y In this case there exist x; y 2 Z such that x1x + x2y = 1 and the matrix has determinant x2 x 1. We can find an explicit completion of (x1; x2) using the Euclidean Algorithm in the following manner: Assume for simplicity that x1; x2 > 0 and x1 Ã> x2. Then!à by division! algorithmà ! x1 = x2q + r, where 1 ¡q x1 r q is the quotient and r is the remainder. Then = . It follows by iterating 0 1 x x 2 2 à ! 1 q the above procedure that we can get a matrix ® which is a product of matrices of the form , 0 1 à ! à ! à ! à ! à ! 1 0 x1 1 1 x1 where q; q0 2 Z such that ® = . Then ®¡1 = and ® 2 0 q 1 x2 0 0 x2 SL2(Z). This example motivates to define En(A). Definition 1.2 — Let A be a commutative ring with identity. Let Eij(¸); i 6= j be the n £ n matrix th in SLn(A) which has 1 as its diagonal entries, ¸ as (i; j) entry and rest of entries are zero. Then the subgroup En(A) generated by Eij(¸); i 6= j of SLn(A) is called elementary group and its elements are called elementary matrices. n For any Euclidean domain A, any unimodular row (a1; a2; ¢ ¢ ¢ ; an) 2 A can be completed to an En(A) En(A) elementary matrix, where n ¸ 2 i.e. (a1; a2; ¢ ¢ ¢ ; an) » (1; 0; ¢ ¢ ¢ ; 0), where » denotes the induced action of En(A) on unimodular rows. En(A) Note that En(A) ½ SLn(A) ½ GLn(A), hence (a1; a2; ¢ ¢ ¢ ; an) » (1; 0; ¢ ¢ ¢ ; 0) ) (a1; a2; ¢ ¢ ¢ ; SLn(A) GLn(A) an) » (1; 0; ¢ ¢ ¢ ; 0) ) (a1; a2; ¢ ¢ ¢ ; an) » (1; 0; ¢ ¢ ¢ ; 0). Since A = k[X] (where k is a field) is En(A) n a Euclidean domain, (a1; a2; ¢ ¢ ¢ ; an) » (1; 0; ¢ ¢ ¢ ; 0) for any unimodular row (a1; a2, ¢ ¢ ¢ ; an) 2 A . n Proposition 1.3 — Let A be a ring and (a1; a2, ¢ ¢ ¢ ; an) 2 A be a unimodular row. Then (a1 + n ¸Xa2; a2; ¢ ¢ ¢ ; an) 2 A[X] is a unimodular row over A[X], for every ¸ 2 A. n n PROOF : Since (a1; a2; ¢ ¢ ¢ ; an) 2 A is a unimodular, there exists (b1; b2; ¢ ¢ ¢ ; bn) 2 A such that CONNECTION BETWEEN COMMUTATIVE ALGEBRA AND TOPOLOGY 473 Pn i=1 aibi = 1. Take c1 = b1; c2 = b2 ¡ b1¸X; c3 = b3; ¢ ¢ ¢ ; cn = bn, thus we have c1(a1 + ¸Xa2) + Pn i=2 aici = 1. Therefore (a1 + ¸Xa2; a2; ¢ ¢ ¢ ; an) is a unimodular row over A[X]. 2 Qr From Proposition 1.3, we can prove that for any matrix σ = i=1 Eij(¸) 2 En(A), (a1; a2; ¢ ¢ ¢ ; an) n Qr σ(X) 2 A[X] is a unimodular over A[X], where σ(X) = i=1 Eij(¸X). 2. TOPOLOGICAL FACT Lemma 2.1 — Let T be a topological space. Then any continuous map f : T ¡! T is homotopic to itself. Lemma 2.2 ([6], Exercise 1, page 325) — Let T;T1 and T2 be topological spaces. Suppose maps 0 0 0 0 h; h : T ¡! T1 and k; k : T1 ¡! T2 are homotopic. Then k ± h and k ± h are homotopic. Lemma 2.3 ([6]) — Let T;T1 and T2 be topological spaces. Suppose k : T ¡! T1 is a continuous 0 map and F is a homotopy between maps f; f : T2 ¡! T , then k ± F is a homotopy between the maps k ± f and k ± f 0. Definition 2.4 — Let f : T ¡! T 0 be a continuous map. We say that f is a homotopy equivalence if 0 0 there exists a continuous map g : T ¡! T such that f ± g is homotopic to the identity map IT 0 on T 0 and g ± f is homotopic to the identity map IT on T . Two spaces T and T are said to be homotopically equivalent or of the same homotopy type if there exists a homotopy equivalence from one to the other. Theorem 2.5 ([3]) — If two spaces T and T 0 are homotopically equivalent and any continuous map from T to Sn¡1 is homotopic to a constant map, then any continuous map from T 0 to Sn¡1 is homotopic to a constant map. Theorem 2.6 ([3]) — Let T be a simplicial complex of dimension · r and T 0 be a closed subcomplex of T . Suppose f : T 0 ¡! Sr is a continuous map, Then f can be extended to a continuous map f 0 : T ¡! Sr such that f 0jT 0 = f. This topological fact can be proved by induction. One chooses a vertex of the complex T that does not belong to T 0 and chooses an arbitrary extension of f. Then one extends f linearly to the edge of T that does not belong to T 0, then to the two faces etc. Theorem 2.7 ([3]) — Let T be a simplicial complex of dimension n and T 0 be a closed subcomplex of T . Then any continuous map f : T 0 ¡! Rn ¡ f0g can be extended to a continuous map f 0 : T ¡! Rn such that f 0¡1(f0g) is a finite set. Theorem 2.8 ([9] Theorem 4.4, page 153) — Let T be a separable metric space and T 0 be a closed subspace of T . Suppose two maps f and g from T 0 ¡! Sn are homotopic. If there exists an extension f 0 : T ¡! Sn of f, then there exists an extension g0 : T ¡! Sn of g such that f 0 and g0 are homotopic. 474 SUMIT KUMAR UPADHYAY et al. m Let I be an ideal of k[X1;X2; ¢ ¢ ¢ ;Xm]. Then V (I) = f(x1, x2, ¢ ¢ ¢ ; xm) 2 k j f(x1; x2; ¢ ¢ ¢ ; xm) = 0, for every f 2 Ig. By Hilbert basis theorem, every ideal of k[X1;X2; ¢ ¢ ¢ , Xm] is finitely generated, so V (I) is the set of common zeros of finitely many polynomials. 2 2 2 1 Example 2.9 : Let A = R[X1;X2] and I = (X1 + X2 ¡ 1) ½ A be an ideal , then V (I) \ R = S (real sphere). Since V (I) \ Rm is the set of common zeros of finitely many polynomials, it is a closed set in the m usual Euclidean topology in R , where I be an ideal of R[X1;X2; ¢ ¢ ¢ ;Xm]. More generally for any field k, there exists a topology on km, where the subsets of the form V (I) are closed. This topology is called the Zariski topology on km. In topology Tietze extension theorem ([6], Theorem 3.2, page 212) says that “Any continuous map of a closed subset of a normal topological space T into the reals R may be extended to a continuous map of T into R”.